Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return b / gcd(a, b) * a; }
void merge(vector<int> &aa, vector<int> &bb) {
  vector<int> ret;
  int idx1 = 0, idx2 = 0;
  while (ret.size() < 10) {
    if (idx1 == aa.size() && idx2 == bb.size()) break;
    if (idx1 == aa.size())
      ret.push_back(bb[idx2++]);
    else if (idx2 == bb.size())
      ret.push_back(aa[idx1++]);
    else if (aa[idx1] < bb[idx2])
      ret.push_back(aa[idx1++]);
    else
      ret.push_back(bb[idx2++]);
    if (ret.size() > 1 && ret[ret.size() - 1] == ret[ret.size() - 2])
      ret.pop_back();
  }
  aa = ret;
}
const int N = 2e5 + 2;
int n, m, q;
vector<int> g[N];
int c[N];
vector<int> in[N];
int u, v, a;
vector<int> dp[N][22];
int st[N][22];
int par[N];
int dep[N];
void DFS(int src, int p) {
  par[src] = p;
  dep[src] = dep[p] + 1;
  for (int i = 0; i < g[src].size(); ++i) {
    int nxt = g[src][i];
    if (nxt == p) continue;
    DFS(nxt, src);
  }
}
int main() {
  cin >> n >> m >> q;
  for (int i = 0; i < n - 1; ++i) {
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 0; i < m; ++i) {
    scanf("%d", &c[i]);
    in[c[i]].push_back(i + 1);
  }
  for (int i = 0; i < N; ++i) {
    sort(in[i].begin(), in[i].end());
    if (in[i].size() > 10) in[i].resize(10);
  }
  DFS(1, 1);
  st[1][0] = -1;
  for (int i = 2; i <= n; ++i) {
    st[i][0] = par[i];
    if (in[par[i]].size()) {
      for (int j = 0; j < min((int)in[par[i]].size(), 10); ++j)
        dp[i][0].push_back((in[par[i]][j]));
    }
  }
  for (int i = 1; i <= n; ++i) sort(dp[i][0].begin(), dp[i][0].end());
  vector<int> aa, bb;
  for (int k = 1; k < 20; ++k) {
    for (int i = 1; i <= n; ++i) {
      if (st[i][k - 1] != -1) {
        st[i][k] = st[st[i][k - 1]][k - 1];
        if (st[i][k] == -1) continue;
        aa = dp[i][k - 1];
        bb = dp[st[i][k - 1]][k - 1];
        int idx1 = 0, idx2 = 0;
        while (dp[i][k].size() < 10) {
          if (idx1 == aa.size() && idx2 == bb.size()) break;
          if (idx1 == aa.size())
            dp[i][k].push_back(bb[idx2++]);
          else if (idx2 == bb.size())
            dp[i][k].push_back(aa[idx1++]);
          else if (aa[idx1] < bb[idx2])
            dp[i][k].push_back(aa[idx1++]);
          else
            dp[i][k].push_back(bb[idx2++]);
        }
      } else
        st[i][k] = -1;
    }
  }
  while (q--) {
    scanf("%d%d%d", &u, &v, &a);
    if (u == v) {
      int M = min(a, (int)in[u].size());
      printf("%d", M);
      for (int i = 0; i < M; ++i) printf(" %d", in[u][i]);
      puts("");
    } else {
      vector<int> res;
      if (dep[u] < dep[v]) swap(u, v);
      int U = u, V = v;
      for (int k = 20; k >= 0; --k) {
        if (dep[u] - (1 << k) >= dep[v]) {
          merge(res, dp[u][k]);
          u = st[u][k];
        }
      }
      for (int k = 20; k >= 0; --k) {
        if (st[u][k] != st[v][k]) {
          merge(res, dp[u][k]);
          merge(res, dp[v][k]);
          u = st[u][k];
          v = st[v][k];
        }
      }
      merge(res, in[U]);
      merge(res, in[V]);
      if (u != v) merge(res, dp[u][0]);
      int M = min(a, (int)res.size());
      printf("%d", M);
      for (int i = 0; i < M; ++i) printf(" %d", res[i]);
      puts("");
    }
  }
}